Finite Gibbs free-energy and entropy optimization

By [F1], write $H={\sum }_j{E}_j\mathrm{\mid }{v}_j⟩⟨v_j\mathrm{\mid }$ with real $E_j$. Then $e^{-\beta H}$ has strictly positive eigenvalues $e^{-\beta E_j}$, so $Z={\sum }_j{e}^{-\beta E_j}>0$, $\sigma$ is faithful, and $\log \sigma =-\beta H-(\log Z)I$.

By [F2] and step 1.1, $D(\rho \mathrm{\parallel }\sigma )=\mathrm{Tr}(\rho \log \rho )+\beta \mathrm{Tr}(\rho H)+\log Z=\beta (\mathrm{Tr}(\rho H)-TS(\rho )-F_{\beta })\mathrm{.}$ All terms are finite because the dimension is finite and $\sigma$ is faithful.

For the inequality and its equality case, diagonalize $\rho ={\sum }_i{w}_i\mathrm{\mid }{u}_i⟩⟨u_i\mathrm{\mid }$ on its support and let $q_j=e^{-\beta E_j}\mathrm{/}Z>0$ be the eigenvalues of $\sigma$ in the $v_j$ basis. Set $r_i={\sum }_j\mathrm{\mid }⟨u_i,v_j⟩{\mathrm{\mid }}^2{q}_j>0$. The overlap weights in each row sum to one, so concavity of $\log$ gives ${\sum }_j\mathrm{\mid }⟨u_i,v_j⟩{\mathrm{\mid }}^2\log q_j\le \log r_i$, with equality only when all $q_j$ having positive overlap in that row are equal. Also $-\log (r_i\mathrm{/}{w}_i)\ge 1-r_i\mathrm{/}{w}_i$, with equality only at $r_i=w_i$. Consequently $D(\rho \mathrm{\parallel }\sigma )\ge {\sum }_{i:w_i>0}{w}_i\log (w_i\mathrm{/}{r}_i)\ge {\sum }_{i:w_i>0}(w_i-r_i)\ge 0,$ because ${\sum }_i{w}_i=1$ and ${\sum }_{i:w_i>0}{r}_i=\mathrm{Tr}(\sigma P_{\mathrm{supp}\rho })\le 1$. Together with step 2.1 this proves the claimed free-energy inequality. If $D=0$, all inequalities are equalities. Strictness of the scalar inequality gives $r_i=w_i$; equality in strict concavity of $\log$ means $u_i$ lies in one eigenspace of $\sigma$, so $\sigma u_i=w_i{u}_i$. The sum of the $r_i$ is then one, so $\mathrm{Tr}(\sigma (I-P_{\mathrm{supp}\rho }))=0$. In an orthonormal basis of that finite complement every diagonal quadratic form of $\sigma$ is nonnegative and their sum is zero, so each vanishes. Its positive square root therefore annihilates that basis and the entire complement. For every $v$ in that orthogonal complement, this gives $\mathrm{\parallel }{\sigma }^{1\mathrm{/}2}v{\mathrm{\parallel }}^2=⟨\sigma v,v⟩=0$, hence $\sigma v=0$; therefore $\sigma$ has no support outside $\mathrm{supp}\rho$. Together with $\sigma u_i=w_i{u}_i$, this proves $\rho =\sigma$. Conversely $\rho =\sigma$ gives $D=0$.

For $\rho =\sigma$, step 1.1 gives $S(\sigma )=-k_B\mathrm{Tr}(\sigma \log \sigma )=k_B(\log Z+\beta \mathrm{Tr}(\sigma H))$. Substitution of $T=(k_B\beta {)}^{-1}$ and $F_{\beta }=-{\beta }^{-1}\log Z$ gives $F_{\beta }=\mathrm{Tr}(\sigma H)-TS(\sigma )$, as stated.